`{:("COLUMN - I","COLUMN - II"),("A) Infinite plane sheet of charge","p)"0),("B) Infinite plane sheet of uniform thickness","q)"(p)/(2epsilon_(0))),("C)Non - conducting charged solid sphere of radius R at its surface","r)"(Rp)/(3epsilon_(0))),("D) Non - conducting charge solid sphere of radius R at its centre","s)"(p)/(epsilon_(0))):}`

Answer

To solve the problem of matching the items in Column I with those in Column II, we will analyze each case step by step.

### Step 1: Analyze the infinite plane sheet of charge (A)

- For an infinite plane sheet of charge with surface charge density \( \sigma \), the electric field \( E \) produced is given by the formula:
\[ E = \frac{\sigma}{2\epsilon_0} \]
- Since the electric field is uniform and directed away from the sheet, the value of \( E \) for this case is \( \frac{p}{2\epsilon_0} \), where \( p \) represents the surface charge density.

**Match: A with q**

### Step 2: Analyze the infinite plane sheet of uniform thickness (B)

- For an infinite plane sheet of uniform thickness, the electric field \( E \) is given by:
\[ E = \frac{\sigma}{\epsilon_0} \]
- Here, the electric field is twice that of a single sheet because there are two surfaces contributing to the electric field.

**Match: B with s**

### Step 3: Analyze the non-conducting charged solid sphere of radius R at its surface (C)

- For a non-conducting charged solid sphere of radius \( R \), the electric field at the surface is derived from Gauss's law:
\[ E = \frac{Q}{4\pi R^2 \epsilon_0} \]
- The total charge \( Q \) can be expressed in terms of the volume charge density \( p \):
\[ Q = p \cdot \frac{4}{3}\pi R^3 \]
- Substituting \( Q \) into the electric field equation gives:
\[ E = \frac{p \cdot \frac{4}{3}\pi R^3}{4\pi R^2 \epsilon_0} = \frac{pR}{3\epsilon_0} \]

**Match: C with r**

### Step 4: Analyze the non-conducting charged solid sphere of radius R at its center (D)

- For a non-conducting charged solid sphere, the electric field inside the sphere (at the center) is zero. This is due to the symmetry of the charge distribution:
\[ E = 0 \]

**Match: D with 0**

### Final Matches

Now, we can summarize the matches:

- A matches with q: \( \frac{p}{2\epsilon_0} \)
- B matches with s: \( \frac{p}{\epsilon_0} \)
- C matches with r: \( \frac{pR}{3\epsilon_0} \)
- D matches with 0: \( 0 \)

### Summary of Matches:
- A → q
- B → s
- C → r
- D → 0